Proof of A Borel Set Whose Lines in One Coordinate Direction Are Null Is Null

We use the notation of the statement.

The measure-theoretic set-up. Let $Z=\{z\}$ be a one-point set and let $\mathcal{G}=\{\varnothing,Z\}$ and $\delta$ be as in claim 3 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, so that $\mathcal{G}$ is a $\sigma$-algebra on $Z$, $\delta(\varnothing)=0$ and $\delta(Z)=1$ define a measure on it, and $\int_{Z}g\,d\delta=g(z)$ for every $g:Z\to[0,\infty]$. The constant sequence with every term $Z$ has union $Z$ and $\delta(Z)=1<\infty$, so $\delta$ is $\sigma$-finite.

We apply Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ with $l=n$, with the measure space $(Z,\mathcal{G},\delta)$ in the role of $(Y,\mathcal{G},\mu)$, and with the given index $i$. By Lebesgue Measure on $\mathbb{R}^n$ the $\sigma$-algebra $\mathcal{B}_{n}$ of that lemma is $\mathcal{B}(\mathbb{R}^{n})$ and the measure $\lambda_{n}$ of that lemma is Lebesgue measure on it, so no distinction between the two notations is needed. When $n=1$ we use the conventions recorded in that lemma for $l=1$, by which $\mathbb{R}^{l-1}\times Y$ denotes $Y$, $\lambda_{l-1}\otimes\mu$ denotes $\mu$, a pair $(\theta',y)$ denotes $y$, and $\Psi_{1}(t,y)=(t,y)$.

The integral of the indicator. The set $N\times Z$ is a product of a member of $\mathcal{B}(\mathbb{R}^{n})$ with a member of $\mathcal{G}$, hence belongs to the product $\sigma$-algebra $\mathcal{B}(\mathbb{R}^{n})\otimes\mathcal{G}$. By The Integral of an Indicator Function is the Measure of the Set its indicator function $\mathbf{1}_{N\times Z}$ is a nonnegative measurable function on $\mathbb{R}^{n}\times Z$ with

$$\int_{\mathbb{R}^{n}\times Z}\mathbf{1}_{N\times Z}\,d(\lambda_{n}\otimes\delta)=(\lambda_{n}\otimes\delta)(N\times Z)=\lambda_{n}(N)\,\delta(Z)=\lambda_{n}(N),$$

the middle equality by Existence and Uniqueness of the Product Measure, which characterises $\lambda_{n}\otimes\delta$ by its values on such products.

By claim 3 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$,

$$\lambda_{n}(N)=\int_{\mathbb{R}^{n-1}\times Z}\Bigl(\int_{\mathbb{R}}\mathbf{1}_{N\times Z}\bigl(\Psi_{i}(t,(\theta',z))\bigr)\,d\lambda(t)\Bigr)\,d(\lambda_{n-1}\otimes\delta)(\theta',z),$$

and for every $(\theta',z)$ the inner integrand is a $\mathcal{B}(\mathbb{R})$-measurable function of $t$.

The inner integral vanishes. Fix $(\theta',z)$ with $\theta'=(\theta_{1},\dots,\theta_{n-1})\in\mathbb{R}^{n-1}$. By the description of the insertion map in claim 2 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$,

$$\Psi_{i}\bigl(t,(\theta',z)\bigr)=\bigl((\theta_{1},\dots,\theta_{i-1},t,\theta_{i},\dots,\theta_{n-1}),\,z\bigr),$$

so $\mathbf{1}_{N\times Z}(\Psi_{i}(t,(\theta',z)))=\mathbf{1}_{T}(t)$, where

$$T=\{\,t\in\mathbb{R}\;:\;(\theta_{1},\dots,\theta_{i-1},t,\theta_{i},\dots,\theta_{n-1})\in N\,\}.$$

Put $y_{0}=(\theta_{1},\dots,\theta_{i-1},0,\theta_{i},\dots,\theta_{n-1})$. For $t\in\mathbb{R}$ the point $y_{0}+te_{i}$ has $i$th coordinate $0+t=t$ and, for $p\neq i$, the same $p$th coordinate as $y_{0}$, because the corresponding coordinate of $e_{i}$ is $0$; hence $y_{0}+te_{i}=(\theta_{1},\dots,\theta_{i-1},t,\theta_{i},\dots,\theta_{n-1})$ and

$$T=\{\,t\in\mathbb{R}\;:\;y_{0}+te_{i}\in N\,\}.$$

By hypothesis $T$ is $\lambda_{1}$-null, so by Null Set of a Measure there is a Borel set $B\subseteq\mathbb{R}$ with $T\subseteq B$ and $\lambda(B)=0$. The function $\mathbf{1}_{T}$ is $\mathcal{B}(\mathbb{R})$-measurable by the previous paragraph, and $T$ is the set where it takes the value $1$, so $T$ is a Borel subset of $\mathbb{R}$; hence claim 2 of Basic Properties of a Measure gives $\lambda(T)\le\lambda(B)=0$. By The Integral of an Indicator Function is the Measure of the Set,

$$\int_{\mathbb{R}}\mathbf{1}_{T}\,d\lambda=\lambda(T)=0 .$$

Conclusion. The inner integral vanishes for every $(\theta',z)$, so the outer integrand is the function taking the value $0$ everywhere on $\mathbb{R}^{n-1}\times Z$; that function is $\mathbf{1}_{\varnothing}$, so by The Integral of an Indicator Function is the Measure of the Set its integral is $(\lambda_{n-1}\otimes\delta)(\varnothing)=0$. Therefore $\lambda_{n}(N)=0$. Since $N\in\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(N)=0$, taking $B=N$ in Null Set of a Measure shows that $N$ is null.